Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(e)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e)

I. Find the I.F. of the following differential equations by transforming them into a linear form.

Question 1.
x\(\frac{dy}{dx}\) – y = 2x² sec² 2x
Solution:
x\(\frac{dy}{dx}\) – y = 2x² sec² 2x
\(\frac{dy}{dx}\) – \(\frac{1}{x}\) . y = 2x. sec² 2x
This is linear in x
I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log x} = e^{log 1/x} = \(\frac{1}{x}\)

Question 2.
y\(\frac{dy}{dx}\) – x = 2y³
Solution:
y\(\frac{dy}{dx}\) – x = 2y³
\(\frac{dy}{dx}\) – \(\frac{1}{y}\).x = 2y²
I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log y} = e^{log 1/y} = \(\frac{1}{y}\)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}\) + y tan x = cos³ x
Solution:

Solution is 2y = x cos x + sin x. cos² x + c. cos x

Question 2.
\(\frac{dy}{dx}\) + y sec x = tan x
Solution:
I.F. = e^{∫sec x dx} = e^{log(secx + tan x)} = sec x + tan x
y. (sec x + tan x)
= ∫tan x (sec x + tan x)dx
= ∫(sec x. tan x + tan² x)dx
= ∫(sec x tan x + sec² x – 1)dx
Solution is
y(sec x + tan x) = sec x + tan x – x + c

Question 3.
\(\frac{dy}{dx}\) – y tan x = e sec x. dx
Solution:
I.F. = e^{-∫tan x dx} = e^{log cos x} = cos x
y. cos x = ∫e^x.sec x. cos x dx = ∫ e^x dx
= e^x + c

Question 4.
x\(\frac{dy}{dx}\) + 2y = log x.
Solution:
I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²
Solution is

Question 5.
(1 + x²)\(\frac{dy}{dx}\) + y = e^{tan-1 x}
Solution:

Question 6.
\(\frac{dy}{dx}+\frac{2y}{x}\) = 2x².
Solution:
I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²
y. x² = ∫2x⁴ dx = \(\frac{2x^5}{5}\) + c

Question 7.
\(\frac{dy}{dx}+\frac{4x}{1+x^2}y=\frac{1}{(1+x^2)^2}\)
Solution:

Question 8.
x\(\frac{dy}{dx}\) + y = (1 + x)e^x
Solution:

Question 9.
\(\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{1+x^2}{1+x^3}\).
Solution:

Question 10.
\(\frac{dy}{dx}\) – y = -2e^-x.
Solution:
I.F = e^∫-dx = e^-x
y. e^-x = -2∫e^-2x dx = e^-2x + c
y = e^-x + c. e^x

Question 11.
(1 + x²)\(\frac{dy}{dx}\) + y = tan^-1 x.
Solution:

Put t = tan^-1 x so that dt = \(\frac{dx}{1+x^2}\)
R.H.S = ∫t. e^tdt = t. e^t – ∫e^tdt
= t. e^t – e^t = e^t(t – 1)
Solution is y. e^{tan-1 x} = e^{tan-1 x} (tan^-11 x – 1) + c
y = tan^-1 x – 1 + c. e^{-tan-1 x}

Question 12.
\(\frac{dy}{dx}\) + y tan x = sin x.
Solution:
I.F. =e^{∫tan x dx} = e^{log sec x} = sec x
y. sec x = ∫ sin x. sec x dx = ∫tan x dx
= log sec x + c

III. Solve the following differential equations.

Question 1.
cos x. \(\frac{dy}{dx}\) + y sin x = sec² x
Solution:
\(\frac{dy}{dx}\) + tan x. y = sec³ x
I.F. = e^{∫tan x dx} = e^{log sec x} = sec x
y. sec x = ∫sec⁴ x dx = ∫(1 + tan² x) sec² x
dx = tan x + \(\frac{\tan^3x}{3}\) + c

Question 2.
sec x. dy = (y + sin x) dx.
Solution:
\(\frac{dy}{dx}=\frac{y+sin x}{sec x}\) = y cos x + sin x. cos x
\(\frac{dy}{dx}\) – y. cos x = sin x. cos x
I.F. = e^{-∫cos x dx} = e^{– sin x}
= y. e^{-sin x} = ∫ e^{-sin x}. sin x. cos x dx ……….. (1)
Consider ∫e^{-sin x}. sin x. cos x dx
t = – sin x ⇒ dt = -cos x dx
∫e^{-sin x}. sin x. cos x dx = + ∫e^t t dt
= t. e^t – e^t + c.
= e^{-sin x} (- sin x – 1) + c
y. e^{-sin x} = – e^{-sin x} (sin x + 1) + c
or y = – (sin x + 1) + c. e^{sin x}.

Question 3.
x log x.\(\frac{dy}{dx}\) + y = 2 log x.
Solution:

Question 4.
(x + y + 1)\(\frac{dy}{dx}\) = 1.
Solution:
\(\frac{dy}{dx}\) = x + y + 1
\(\frac{dy}{dx}\) – x = y + 1
I.F. = e^∫-dy e^-y
x . e^-y = ∫e^-y (y + 1) dy
= -(y + 1). e^-y + ∫e^-y. dy
= -(y + 1) e^-y – e^-y
= -(y + 2) e^-y+ c
x = -(y + 2) + c. e^-y

Question 5.
x(x – 1)\(\frac{dy}{dx}\) – y = x³(x – 1)³
Solution:
\(\frac{dy}{dx}-\frac{1}{x(x-1)}\)y = x²(x – 1)²

Question 6.
(x + 2y³)\(\frac{dy}{dx}\) = y
Solution:

Solution is x = y(y² + c)

Question 7.
(1 – x²)\(\frac{dy}{dx}\) + 2xy = x\(\sqrt{1-x^2}\)
Solution:

Question 8.
x(x – 1)\(\frac{dy}{dx}\) – (x – 2)y = x³(2x – 1)
Solution:


Solution is y(x – 1) = x²(x² – x + c)

Question 9.
\(\frac{dy}{dx}\)(x²y³ + xy) = 1
Solution:
\(\frac{dy}{dx}\) = xy + x²y³
This is Bernoulli’s equation

Question 10.
\(\frac{dy}{dx}\) + x sin 2y = x³ cos² y
Solution:

Question 11.
y² + (1 – \(\frac{1}{y}\)).\(\frac{dy}{dx}\) = 0
Solution:


Solution xy = 1 + y + cy.e^1/y